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Aqueous Equilibria, Part Two

AP Chemistry. Aqueous Equilibria, Part Two. CH 3 COOH . NaCH 3 COO . CH 3 COOH CH 3 COO – + H +. greatly [CH 3 COO – ]. shift. [H + ] and pH. sodium acetate. The Common-Ion Effect. Consider a mixture of a “weak” and a “common-ion,” soluble salt. .

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Aqueous Equilibria, Part Two

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  1. AP Chemistry Aqueous Equilibria, Part Two

  2. CH3COOH NaCH3COO CH3COOH CH3COO– + H+ greatly [CH3COO–]. shift [H+] and pH sodium acetate The Common-Ion Effect Consider a mixture of a “weak” and a “common-ion,” soluble salt. Since NaCH3COO is strong, adding it to the solution… By Le Chatelier… The result is that…

  3. HBrO H+ + BrO– This illustrates the common-ion effect: “The dissociation of a ‘weak’ DECREASES when a strong, common-ion salt is added to the solution.” For 100 HBrOs in soln… 90 10 10 If we add a “strong,” BrO– salt (like KBrO), the amount of BrO– goes up, and the amount of HBrO that dissociates goes down (by Le Chatelier).

  4. x2 0.10x + x2 = 4.5 x 10–4 = 0.085 – x 0.085 – x Find the pH of a solution containing 0.085 M nitrous acid (Ka = 4.5 x 10–4) and 0.10 M potassium nitrite. HNO2 H+ + NO2– – x + x + x 0.085 0 0.10 X Ka = 4.5 x 10–4 X So… x = [H+] = 3.825 x 10–4 M pH = 3.42 Find the pH of 0.085 M nitrous acid, on its own. HNO2 DID dissociate less when the common-ion salt was present. X So… x = [H+] = 6.185 x 10–3 M pH = 2.21

  5. Buffered Solutions (“buffers”) solns based on the common-ion effect -- they resist DpH -- -- a mixture of… a “weak” (either acid or base) and a common-ion salt NH3 and NH4Cl HOAc and NaOAc e.g., HF and KF HBrO2 and Ca(BrO2)2 buffer capacity: the amount of acid or base the buffer can “neutralize” before the pH begins to change appreciably greater amounts of acid/base AND common-ion salt -- buffer capacity increases with…

  6. 0.10x + x2 1.4 x 10–4 = 0.12 – x “You betcha.” Find the pH of a buffer that is 0.12 M lactic acid, HC3H5O3 (Ka = 1.4 x 10–4) and 0.10 M sodium lactate. (NaC3H5O3) HC3H5O3 H+ + C3H5O3– + x + x – x 0.12 0 0.10 “Shortcut OK?” X X 1.68 x 10–4 M pH = 3.77 x = [H+] =

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